Generalized Eigenspace Decomposition

Every vector space decomposes into generalized eigenspaces, providing a bridge between Jordan form and abstract operator theory. This decomposition generalizes the diagonalization perspective.

TheoremPrimary Decomposition Theorem

Let $A$ be an $n \times n$ matrix with distinct eigenvalues $\lambda_1, \ldots, \lambda_k$ . Then: $\mathbb{C}^n = G_{\lambda_1} \oplus G_{\lambda_2} \oplus \cdots \oplus G_{\lambda_k}$

where $G_{\lambda_i} = \ker((A - \lambda_i I)^{n})$ is the generalized eigenspace.

Moreover:

Each $G_{\lambda_i}$ is $A$ -invariant
$\dim(G_{\lambda_i})$ equals the algebraic multiplicity of $\lambda_i$
The restriction $A|_{G_{\lambda_i}}$ has only eigenvalue $\lambda_i$

TheoremFitting Lemma

For linear operator $T$ on finite-dimensional space $V$ , if $p(T) = 0$ for polynomial $p = p_1p_2$ with $\gcd(p_1, p_2) = 1$ , then: $V = \ker(p_1(T)) \oplus \ker(p_2(T))$

This decomposition is $T$ -invariant: both subspaces are preserved by $T$ .

Applied to minimal polynomial $m_A(\lambda) = \prod(\lambda - \lambda_i)^{k_i}$ , this gives the primary decomposition.

ExampleDecomposition into Generalized Eigenspaces

For $A = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ :

Eigenvalues: $\lambda_1 = 2$ , $\lambda_2 = 3$

$G_2 = \ker((A-2I)^2) = \text{span}\left\{\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\right\}$ (dimension 2)

$G_3 = \ker(A-3I) = \text{span}\left\{\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\right\}$ (dimension 1)

Indeed $\mathbb{C}^3 = G_2 \oplus G_3$ .

TheoremInvariant Subspace Decomposition

The Jordan form corresponds to choosing a basis that:

Respects the primary decomposition $V = \bigoplus G_{\lambda_i}$
Within each $G_{\lambda_i}$ , organizes vectors into Jordan chains

This two-level structure (decompose by eigenvalue, then by chain) explains the block diagonal Jordan matrix.

TheoremCommuting Operators

If linear operators $S$ and $T$ commute ( $ST = TS$ ), then:

Eigenspaces of $T$ are $S$ -invariant
Generalized eigenspaces of $T$ are $S$ -invariant
$S$ and $T$ can be simultaneously put in upper triangular form (though not necessarily Jordan form)

Remark

The generalized eigenspace decomposition reveals Jordan form as a refinement of spectral theory: while diagonalizable matrices split as direct sums of one-dimensional eigenspaces, general matrices split as direct sums of generalized eigenspaces, within which Jordan blocks organize the cyclic structure. This perspective connects Jordan theory to module theory and representation theory.